Question

Two point charges, -62.5 microcoulombs (placed on the left side of the positive charge) and +12.5 microcoulombs, fixed on the x-axis, are seperated by a distance of 25.0 cm from each other. At what points along the x-axis is the electric potential equal to zero? (A well-labeled sketch showing these points is required Find the voltage at a point on the x-axis, 15.0 cm to the left of the negative charge.

Answer 1

1) from fig-1, let at point P at a distance x from -62.5uc charge potential is zero.

V1 = potential at P due to q1 = (1/4*pie*E₀) * (q1/r1 )

                                         = 9*10⁹* ( - 62.5*10^-6) / x

V2 = (1/4*pie*E₀)* (q2/r2)

    = 9*10⁹ * 12.5*10^-6 / (25-x)

net potential at P, V = V1 +V2                                          ( net potential V = 0)

0 = (- 9*10⁹*62.5*10^-6 / x ) + ( 9*10⁹*12.5*10^-6 / 25-x)

on solving this, x= 20.83cm

ie potential is zero at a distance 20.83 cm from -62.5uc charge or at a distance 4.17cm from 12.5 uc charge.

2) from fig 2 , we have to calculate voltage( potential) at point P

V1 = voltage due to q1 = kq1/ 15                          ( k = 1/4*pie*E₀ = 9*10⁹ )

V2 = k*q2 / 40 cm

net voltage, V = V1 +V2

                    =[ 9*10⁹* ( - 62.5*10^-6) / 15*10^-2 ] + [ 9*10⁹ *12.5*10^-6 / 40*10^-2 ]

on solving this,   V = - 3.465*10⁶ volt